Java 解法, 执行用时: 31ms, 内存消耗: 12244K, 提交时间: 2021-05-28 19:50:29

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s1 = sc.next();
        String s2 = sc.next();
        char[] a1 = s1.toCharArray();
        char[] a2 = s2.toCharArray();
        int len = a1.length;
        int count = 0;
        for(int i = 0; i< len ; i++){
            if( a1[i] != a2[i])count++;
        }
        if(count <= 2)
            System.out.println("yeS");
        else
            System.out.println("N0");
    }
}

C(clang11) 解法, 执行用时: 2ms, 内存消耗: 368K, 提交时间: 2021-04-20 13:40:26

#include <stdio.h>
int main()
{
    char s1[1005],s2[1005];
    scanf("%s%s",s1,s2);
    int sum = 0;
    for (int i = 0;i < strlen(s1);i ++){
        if (s1[i] != s2[i]){
            sum ++;
        }
    }
    if (sum == 2 || sum == 0){
        printf("yeS");
    } else {
        printf("N0");
    }
    return 0;
}

C++ 解法, 执行用时: 6ms, 内存消耗: 380K, 提交时间: 2021-05-28 17:10:47

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
	string a,b;
	cin>>a>>b;
	int x=0;
	for(int i=0;i<a.length() ;i++){
		if(a[i]!=b[i]) x++;
	}
	if(x>2) cout<<"N0";
	else cout<<"yeS";
	return 0;
}

Python3(3.9) 解法, 执行用时: 33ms, 内存消耗: 6884K, 提交时间: 2021-04-20 15:45:16

s1 =list(input())
s2 =list(input())
num = 0
for i in range(len(s1)):
    if s1[i]!=s2[i]:
        num += 1
if num==0 or num==2:
    print('yeS')
else :
    print('N0')